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प्रश्न
Find \[\left( x + 1 \right)^6 + \left( x - 1 \right)^6\] . Hence, or otherwise evaluate \[\left( \sqrt{2} + 1 \right)^6 + \sqrt{2} - 1^6\] .
उत्तर
The expression \[\left( x + 1 \right)^6 + \left( x - 1 \right)^6\] can be written as \[(x + 1 )^6 + (x - 1 )^6 \]
\[ = 2[ ^{6}{}{C}_0 x^6 +^{6}{}{C}_2 x^4 + ^{6}{}{C}_4 x^2 + ^{6}{}{C}_6 x^0 ]\]
\[ = 2[ x^6 + 15 x^4 + 15 x^2 + 1]\]
By taking \[x = \sqrt{2}\] , we get:
\[(\sqrt{2} + 1 )^6 + (\sqrt{2} - 1 )^6 = 2[(\sqrt{2} )^6 + 15(\sqrt{2} )^4 + 15(\sqrt{2} )^2 + 1]\]
\[ = 2 \times (8 + 60 + 30 + 1)\]
\[ = 198\]
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\[= ^{5}{}{C}_0 (2x )^5 (3y )^0 +^{5}{}{C}_1 (2x )^4 (3y )^1 + ^{5}{}{C}_2 (2x )^3 (3y )^2 + ^{5}{}{C}_3 (2x )^2 (3y )^3 + ^{5}{}{C}_4 (2x )^1 (3y )^4 +^{5}{}{C}_5 (2x )^0 (3y )^5\]
\[= 32 x^5 + 5 \times 16 x^4 \times 3y + 10 \times 8 x^3 \times 9 y^2 + 10 \times 4 x^2 \times 27 y^3 + 5 \times 2x \times 81 y^4 + 243 y^5 \]
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